SPOJ SUBLEX 7258. Lexicographical Substring Search

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看起来像是普通的SAM+dfs...但SPOJ太慢了......倒腾了一个晚上不是WA 就是RE .....

最后换SA写了......

Lexicographical Substring Search Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status]   Description Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string S and asked him Q questions of the form: If all distinct substrings of string S were sorted lexicographically, which one will be the K-th smallest? After knowing the huge number of questions Kinan will ask, Daniel figured out that he can‘t do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given S will answer Kinan‘s questions. Example: S = "aaa" (without quotes) substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be: "a", "aa", "aaa".   Input In the first line there is Kinan‘s string S (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer Q (Q <= 500) , the number of questions Daniel will be asked. In the next Q lines a single integer K is given (0 < K < 2^31). Output Output consists of Q lines, the i-th contains a string which is the answer to the i-th asked question. Example Input: aaa 2 2 3 Output: aa aaa Source Own Problem [Submit]   [Go Back]   [Status]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=99000;

int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn];
int *x,*y,ans[maxn];
char str[maxn];

bool cmp(int*r,int a,int b,int l,int n)
{
    if(r[a]==r[b]&&a+l<n&&b+l<n&&r[a+l]==r[b+l])
        return true;
    return false;
}

void radix_sort(int n,int sz)
{
    for(int i=0;i<sz;i++) c[i]=0;
    for(int i=0;i<n;i++) c[x[y[i]]]++;
    for(int i=1;i<sz;i++) c[i]+=c[i-1];
    for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
}

void get_sa(char c[],int n,int sz=128)
{
    x=rank,y=rank2;
    for(int i=0;i<n;i++)
        x[i]=c[i],y[i]=i;
    radix_sort(n,sz);
    for(int len=1;len<n;len=len*2)
    {
        int yid=0;
        for(int i=n-len;i<n;i++)
            y[yid++]=i;
        for(int i=0;i<n;i++)
            if(sa[i]>=len)
                y[yid++]=sa[i]-len;

        radix_sort(n,sz);

        swap(x,y);
        x[sa[0]]=yid=0;

        for(int i=1;i<n;i++)
            x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n)?yid:++yid;

        sz=yid+1;
        if(sz>=n) break;
    }
    for(int i=0;i<n;i++)
        rank[i]=x[i];
}

void get_h(char str[],int n)
{
    int k=0; h[0]=0;
    for(int i=0;i<n;i++)
    {
        if(rank[i]==0) continue;
        k=max(k-1,0);
        int j=sa[rank[i]-1];
        while(i+k<n&&j+k<n&&str[i+k]==str[j+k])
            k++;
        h[rank[i]]=k;
    }
}

int sum[maxn];

int main()
{
    scanf("%s",str);
    int n=strlen(str);
    get_sa(str,n);
    get_h(str,n);

    for(int i=0;i<n;i++)
    {
        sum[i]=n-sa[i]-h[i];
        if(i-1>=0) sum[i]+=sum[i-1];
    }

    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        int x;
        scanf("%d",&x);
        int low=0,high=n-1,ans,mid;
        while(low<=high)
        {
            mid=(low+high)/2;
            if(sum[mid]>=x)
                ans=mid,high=mid-1;
            else if(sum[mid]<x) low=mid+1;
            else break;
        }
        int t=(ans==0)?0:sum[ans-1];
        for(int i=0;i<h[ans];i++)
            putchar(str[sa[ans]+i]);
        for(int i=0;i<x-t;i++)
            putchar(str[sa[ans]+h[ans]+i]);
        putchar(10);
    }
    return 0;
}

SPOJ SUBLEX 7258. Lexicographical Substring Search,布布扣,bubuko.com

SPOJ SUBLEX 7258. Lexicographical Substring Search

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原文地址：http://blog.csdn.net/ck_boss/article/details/37053289